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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Oct 2008 18:17:00 IST
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can anyone help me solve this question?
limit of (-lnx)^x when x approach 0 from the right [positive limit]
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i wan go iit as a exchange student next year(2009) |
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4 Oct 2008 19:33:41 IST
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always..write by basic method..
- ln(h) ^ h..
-lnh / 1 / h ...
tjis is infinity by infinity form..apply lh..
- 1 / h / -1 /h^2
= h...
or simply 0 (+)..
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Don't Dream ..Do the dream...
Rock On ....
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4 Oct 2008 22:09:18 IST
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can u explain in more detail,i still cant get it
the question is (-lnx)^x
is power x,not x(-lnx)
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i wan go iit as a exchange student next year(2009) |
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4 Oct 2008 22:12:44 IST
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i am getting the answer as 1..is it right
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4 Oct 2008 22:23:47 IST
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yea,is 1!can u show me how did u work it out?please
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i wan go iit as a exchange student next year(2009) |
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4 Oct 2008 23:18:20 IST
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let l = lim x->0 (-lnx)^x
taking log
log l = lim x->0 - (x) (lnx)
convert the right hand side onto 0/0 form
log l = lim x->0 - (lnx) / (1/x)
applting L hospital rule
log l = lim x->0 - (1/x) / ( -1/x2)
log l = lim x->0 x
log l = 0
l = e0
l = 1
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5 Oct 2008 20:39:16 IST
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oh !! sorry frind i took tht ^ sign..as * ...don't give heed to it...
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Don't Dream ..Do the dream...
Rock On ....
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